# Quadratic Equations - Hemant Malhotra

• Find maximum number of real roots(+ve ,-ve) for the equations x^3+6x^2+11x-6=0

The max number of +ve real roots of a polynomial f(x)=0 is d number of changes of signs .
the max number of -ve real roots of f(x)=0 is number of changes of sign in f(-x)=0
so f(x)=x^3+6x^2+11x-6
positive real roots = + + + - so 1 sign change so max number of +ve real roots=1
for negative real roots f(-x)=-x^3+6x^2-11x-6
sign changes are as - + - - : 2 changes so 2 negative real roots max

this rule is known as Descarte's rule of sign. This will give u only max number of real roots .. there is possibility that roots are less than that

Find equation whose roots are reciprocal of the roots of ax^2+bx+c=0
a) cx^2+bx+a
b) bx^2+cx+A=0
c) cx^2+ax+b=0
d) bx^2+ax+c=0

Let we have an equation ax^2+bx+C=0and roots of this is p and q now find equation whose roots are reciprocal to roots of this given equation
-- reciprocal so just replace x by 1/x
a(1/x)^2+b(1/x)+c=0 so cx^2+bx+a=0
let in same question we want equation whose roots are cube of give equation so
so just replace x by x^1/3
NOTE_ in square we will replace x by x^1/2 , cube by x^1/3 , in cube root x by x^3 and so on :) ... let we have to find equation whose roots are 2 more than the roots of given equation then replace x by x-2 .. so we have to follow just reverse process.

If sum of two roots of equation x^3-px^2+qx-r is zero. then
a) pq = r
b) qr = p
c) pr = q
d) pqr = 1

sum of roots=a+b+c=p
given a+b=0
so c=p
we know that c is a root of equation so it will satisfy
so p^3-p^3+qp-r=0
so pq=r

If a and b( does not equal to zero) are roots of x^2 + ax + b = 0 find least value of x^2 + ax + b ( x is real)
a) 9/4
b) -9/4
c) -1/4
d) 1/4

a and b are roots of x^2+ax+b
so a+b = -a
so 2a+b=0
now product of roots=a*b=b
b(a-1)=0

case1- a-1=0 so a=1
2a+b=0 so b=-2

case2-
b=0

but this is given that b is not zero
so a=1 and b=-2
so x^2+ax+b ( min)
= -D/4a=4b-a^2/4=-9/4

min value of ax^2+bx+c is -D/4a

If the equation 2x^2+kx-5=0 and x^2-3x-4=0 have one root in common then k
a) -3
b) -27/4
c) 27/4
d) not

let p is common root
2p^2+kp-5=0
p^2-3p-4=0
now solve these equations
p^2/(-4k-15)=p/(-5+8)=1/(-6-k)
so p^2=(4k+15)/(k+6)
and p =-3/(k+6)
so (-3/k+6)^2=4k+15/(k+6)
so k=-3 or -27/4

The number of values of pair (a,b) for which a(x+1)^2+b(-x^2-3x-2)+x+1=0 is an identity in x
a) 0
b) 1
c) 2
d) infinite

IF ANY QUADRATIC EQUATION ax^2+bx+c=0 HAS MORE THAN TWO ROOTS THEN IT BECOMES AN IDENTITY SO IN THAT CASE a=b=c=0

Number of common roots between two equations x^3+3x^2+4x+5=0 and x^3+2x^2+7x+3=0
a) 0
b) 1
c) 2
d) 3

f(x)=x^3+3x^2+4x+5
g(x)=x^3+2x^2+7x+3
f(x)-g(x)
x^2-3x+2=0
x=1,2
( now when u got this root there may be the case that ye value initial equatuion ko satisfy na kare so we should always check this_ these roots are know as false roots ...
put x=1 and x=2 in initial equation also

x^3-6x^2+15x+3=0 has
a) only one positive root
b) 2 +ve and 1 -ve
c) no +ve root
d) not

Roll's theroem
it states that if two value a and b are such that f(a)>0 and
f(b) < 0
then there must be a real root in between a and b
Example -
if f(x)=f(x)=2^x-x^2+1
f(3)=0
f(3.2) < 0
f(4)=2^4-16+1>0
so one root will lie between 3 and 4
now f(-1)=1/2-1+1>0
f(-2)=1/4-4+1
so one root lie between -1 and -2
and f(3)=0 so one root 3
so 3 roots

every equation of an odd degree has at least one real root
x^3-6x^2+15x+3=0 has atleast one real root and max real roots=3
now question
f(x)=x^3-6x^2+15x+3
f'(x) ((differentiation)
3x^2-12x+15=3((( x-2)^2+1))
so f'(x)>0 means function is increasing so it will cut x axis only once so one real root only
so this equation has one real root either positive or negative
now f(0)=3
f(-1) < 0
so one root will lie between o and -1 so negative real root

x^3-ax^2+bx-a=0 has three real roots then which is true
a) a=11
b) a not equal to 1
c) b=1
d) b not equal to 1

x^3-ax^2+bx-a=0
(x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rx)x-pqr
now compare
b=pq+qr+rp
a=p+q+r=pqr
(p+q+r/pqr)=1
sk 1/pq+1/qr+1/rp=1
so pq+qr+rp>3 so b does not equal to 1

root(x+1)-root(x-1)=root(4x-1) find number of real values of x
a) 1
b) 0
c) 2
d) not

when you solve, you will get x=5/4 but that value will not satisfy initial equation so zero roots

Find integral value of a for which x^2 - 2(4a - 1)x + 15a^2 - 2a - 7 > 0 is valid for any x
a) 2
b) 3
c) 4
d) no

if f(x) > 0 for all values of x
then D < 0
4(4a-1)^2-4(15a^2-2k-7) < 0
a^2-6a+8 < 0
2 < a < 4
so integral value is 3

If roots p,q,r in HP of equation x^3 - 3ax^2 + 3bx - c = 0 then
a) q=1/p
b) q = b
c) q = c/b
d) q = b/c

approach 1
p+q+r = 3a
pq + qr + rp = 3b
pqr = c
q = 2pr/p+r
1 = 2c/q / [3b-c/q]
3b = 3c/q
q=c/b

approach 2
roots are in HP
so 1/p,1/q,1/r are the roots of
(1/x)^3-3z(1/x)^2+3b(1/x)-c=0
so -cx^3+3bx^2-3ax+1=0 are in AP
so 1/p+1/q+1/r=3b./c
so 3/q=3b/c
so q=c/b

For all real values of find min and max value of (x^2-3x+4)/(x^2+3x+4)

x^2-3x+4/(x^2+3x+4) = y
now make a quadratic in x
x^2(y-1)+3x(y+1)+4(y-1) = 0
now real x so D > = 0
9(y+1)^2-16(y-1)^2 > = 0
so (7y-1)(y-7) < = 0
so 1/7 < = y < = 7

Number of possible value of integer p for which x^2 + px + 16 = 0 has integral roots
a) 4
b) 6
c) 2
d) not

For roots to be integral Discriminant should be perfect square
D=k^2
a^2-64=k^2
a^2-k^2=64
Now easy ? Answer is 6

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